Summation of Combination Terms MCQ Quiz in मराठी - Objective Question with Answer for Summation of Combination Terms - मोफत PDF डाउनलोड करा

Calculation:

Given,

The equation is \( (1 + x + x^2)^{2n} \)

We are asked to determine the correct relationship between the coefficients \(a_r \) in the expansion of \( (1 + x + x^2)^{2n} \).

The expansion of \( (1 + x + x^2)^{2n} \) is given by:

\( (1 + x + x^2)^{2n} = \sum_{r=0}^{4n} a_r \cdot x^r \)

From the structure of the binomial expansion of \( (1 + x + x^2)^{2n} \), we can observe that the coefficients \(a_r \) follow a symmetry. Specifically, the coefficients on opposite ends of the expansion are equal. That is:

\( a_r = a_{4n - r}, \quad 0 \leq r \leq 4n \)

This symmetry implies that:

\( a_0 = a_{4n} \)

\( a_1 = a_{4n-1} \)

\( a_2 = a_{4n-2} \)

and so on.

Therefore, the correct statement is:

\( a_r = a_{4n - r}, 0 \leq r \leq 4n \)

Hence, the correct answer is Option (4) 

Calculation:

Given,

The equation is \( (1 + x + x^2)^{2n} \)

We need to find the value of \( a_{4n-1} \), which is the coefficient of \( x^{4n-1} \) in the expansion of \( (1 + x + x^2)^{2n} \).

We first replace x by\( \frac{1}{x} \) in the given expression to get the equation:

\( (1 + x + x^2)^{2n} = \sum_{r=0}^{4n} a_r x^r \)

We then modify the expression as follows:

\( (1 + x + x^2)^{2n} = \sum_{r=0}^{4n} a_r x^r = \sum_{r=0}^{4n} a_r x^{4n-r} \)

By comparing the coefficients, we obtain:

\( a_r = a_{4n-r} \)

Thus, the value of \(a_{4n-1} \) is equal to the coefficient of x  in the expansion of \( (1 + x + x^2)^{2n} \), which is \( 2nC_1 \).

Hence, the value of \( a_{4n-1} \) is \( 2n \).

Hence, the correct answer is Option (1) 

Calculation:

Given,

The equation is \( (1 + x + x^2)^{2n} \)

We need to find the value of \( a_2 \), which is the coefficient of \( x^2 \) in the expansion of \( (1 + x + x^2)^{2n} \).

The equation is expanded as follows:

\( (1 + x + x^2)^{2n} = 1 + 2n C_1 (x + x^2) + 2n C_2 (x + x^2)^2 + \dots \)

We are interested in the coefficient of \( x^2 \).

From the binomial expansion, we have:

\( a_2 = 2n C_1 + 2n C_2 \)

\( a_2 = 2n + 1 C_2 \)

Hence, the correct answer is Option (3) 

Given:

Series is \(\sum^n_1{\frac{1}{(x+3)(x+4)}}\)

Concept Used:

\({\frac{1}{(a)(b)}} = {\frac{1}{(a)}} - {\frac{1}{(b)}}, \space where (a < b)\)

Calculation:

\(\sum^n_1{\frac{1}{(x+3)(x+4)}} = {\frac{1}{(1+3)(1+4)}} + {\frac{1}{(2+3)(2+4)}} + ..........+ {\frac{1}{(n+3)(n+4)}}\)

\(\Rightarrow {\frac{1}{4.5}} + {\frac{1}{5.6}} + ..........+ {\frac{1}{(n+3)(n+4)}}\)

\(\Rightarrow {\frac{1}{4}} - {\frac{1}{5}} + {\frac{1}{5}} - {\frac{1}{6}}..........+ {\frac{1}{(n+3)}} - {\frac{1}{(n+4)}}\)

\(\Rightarrow {\frac{1}{4}} - {\frac{1}{(n+4)}}\)

\(\Rightarrow {\frac{n + 4 - 4}{4.(n+4)}}\)

\(\therefore The \space value \space of \space \sum^n_1{\frac{1}{(x+3)(x+4)}} = {\frac{n}{4.(n+4)}}\)